Application of Derivatives 4 Question 15

16. Let f(x) be a polynomial of degree four having extreme values at x=1 and x=2. If limx01+f(x)x2=3, then f(2) is equal to

(a) -8

(b) -4

(c) 0

(d) 4

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Answer:

Correct Answer: 16. (c)

Solution:

  1. PLAN Any function have extreme values (maximum or minimum) at its critical points, where f(x)=2.

Since, the function have extreme values at x=1 and x=2.

f(x)=0 at x=1 and x=2f(1)=0 and f(2)=0

Also, it is given that,

limx01+f(x)x2=31+limx0f(x)x2=3limx0f(x)x2=2

f(x) will be of the form ax4+bx3+2x2.

[f(x) is of four degree polynomial]

 Let f(x)=ax4+bx3+2x2f(x)=4ax3+3bx2+4xf(1)=4a+3b+4=0 and f(2)=32a+12b+8=08a+3b+2=0

On solving Eqs. (i) and (ii),

 we get a=12,b=2f(x)=x422x3+2x2f(2)=816+8=0



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